IPv6 Ignorance
Owen DeLong
owen at delong.com
Wed Sep 19 05:13:59 UTC 2012
I won't dispute that, but let's look at some of the densest uses of it, factoring in the vertical aspects
as well...
Let's assume an 88 story sky scraper 1 city block square (based on an average of 17 city block/mile).
That's 96,465 sq. feet (8,961,918 sq. cm.) total building foot print.
Subtract roughly 1,000,000 sq. cm. for walls, power, elevators, risers, etc leaves us
with 7,961,918 sq. cm. per floor.
Figure in a building that large, you probably need 5 floors for generators, 8 floors for chiller plants,
and another 2 floors or more for other mechanical gives us a total of 73 datacenter floors max.
(Which I would argue is still unrealistic, but what the heck).
Subtract 1/3rds of the datacenter area for PDUs and CRAC units puts us at 5,307,945 sq. cm.
per floor.
FIguring a typical cabinet occupancy area + aisles of 2'x6' (small on the aisles, actually) gives us 12 sq. ft
per cabinet = 11,148 sq. cm. per cabinet so we get roughly 715 cabinets per floor (max) and let's assume
each 1U server holds 1000 virtual hosts at 42 servers per cabinet, that's 30,030 addresses per cabinet.
Multiplied by 75 floors, that's a building total of 2,252,250 total addresses needed. We haven't even
blown out a single /64 (and that's without allowing for the lower address density on routers, core switches,
etc.).
Let's assume we want to give a /64 to each server full of virtual hosts, we're still only taliking about 53,625
/64s, so the whole building can still be addressed within a /48 pretty easily (unless you think you have
more than 12,000 additional point-to-point/other administrative/infrastructure links within the building in
which case, you might need as much as a /47.)
In terms of total addresses per cm, 2,252,250 addresses spread over the building footprint of 8,961,918
sq. cm. is still only 0.25 addresses per sq. cm. so it falls well short of the proposed 2 addresses per
sq. cm.
To even achieve the suggested 2 addresses per sq. cm, you would need to make the building
704 stories tall and still dense-pack every possible sq. foot of the building with datacenter and
you'd have to put these kinds of buildings EVERYWHERE on earth, including over the oceans.
I'm willing to say that based on that math, there are more than enough addresses for virtually any
rational addressing scheme.
Owen
On Sep 18, 2012, at 09:01 , "Beeman, Davis" <Davis.Beeman at integratelecom.com> wrote:
> Orbits may not be important to this calculation, but just doing some quick head math, I believe large skyscrapers could already have close to this concentration of addresses, if you reduce them down to flat earth surface area. The point here is that breaking out the math based on the surface area of the earth is silly, as we do not utilize the surface of the earth in a flat manner...
>
> Davis Beeman
>
>
>> On Mon, Sep 17, 2012 at 11:27:04AM -0700, Owen DeLong wrote:
>>
>>> What technology are you planning to deploy that will consume more than 2 addresses per square cm?
>>
>> Easy. Think volume (as in: orbit), and think um^3 for a functional
>> computers ;)
>
> I meant real-world application.
>
> Orbits are limited due to the required combination of speed and altitude. There are a limited number of achievable altitudes and collision avoidance also creates interesting problems in time-slotting for orbits which are not geostationary.
>
> Geostationary orbits are currently limited to one object per degree of earth surface, and even at 4x that, you could give every satellite a /48 and still not burn through a /32.
>
> Owen
>
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